A student sought my help with a logic problem concerning arrays and loops. The challenge was to determine if there is a 1 in the array with a 2 occurring later in the array. Let’s go through the student’s initial attempt at solving the problem:

The Initial Code

javaCopy codefor (int i = 0; i < nums.length; i++) {
    if (nums[i] == 1 && nums[i] == 2) {
        return true;
    }
}
return false;

The for loop appeared to be correct, iterating through the array’s length and accessing elements properly. However, the conditional check inside the loop had a small mistake.

The Key Component

The problem was that the code checked if any element in the array was both 1 and 2 at the same time, which is impossible. Instead, we need to check for a 1 first and then search for a 2 later in the array. To achieve this, we can use a boolean variable, let’s call it has1, initially set to false. This variable will help us remember if we have found a 1 while iterating through the array.

Corrected Solution

javaCopy codeboolean has1 = false;

for (int i = 0; i < nums.length; i++) {
    if (nums[i] == 1) {
        has1 = true; // We found a 1
    } else if (has1 && nums[i] == 2) {
        return true; // We found a 2 after a 1
    }
}

return false; // 1 and 2 were not found in the required order

Now, we first check for 1, and if the number is not 1, we verify if we have already found a 1 (has1) and if the current number is 2 (nums[i] == 2). This approach helps us find a 2 occurring after a 1 in the array.

Conclusion

Remember, when your program needs to “remember” something, you should use a variable. Consider what information your program needs to retain and what data type can best represent it. In cases where you need a simple switch to indicate existence or absence, boolean variables are your solution.

Happy coding!

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